This post assumes familiarity with some basic concepts in complex analysis, including continuity and entire (everywhere complex-differentiable) functions. This is likely the simplest proof of the theorem (at least, among those that this author has seen), although it stands on the shoulders of a highly nontrivial theorem.

The fundamental theorem of algebra has quite a few number of proofs (enough to fill a book!). In fact, it seems a new tool in mathematics can prove its worth by being able to prove the fundamental theorem in a different way. This series of proofs of the fundamental theorem also highlights how in mathematics there are many many ways to prove a single theorem, and in re-proving an established theorem we introduce new concepts and strategies. And perhaps most of all, we accentuate the unifying beauty of mathematics.

Problem: Let be a non-constant polynomial. Prove has a root in .

Solution:Picard’s Little Theorem states that any entire function whose range omits two distinct points must be constant. (See here for a proof sketch, and other notes)

Suppose to the contrary that has no zero. We claim also does not achieve some number in the set . Indeed, if it achieved all such values, we claim continuity would provide a zero. First, observe that as is unbounded for large enough , we may pick a such that whenever . Then the points such that all lie within the closed disk of radius . Because the set is closed and the complex plane is a complete metric space, we see that the sequence of converges to some , and by continuity .

In other words, no such sequence of can exist, and hence there is some omitted in the range of . As polynomials are entire, Picard’s Little theorem applies, and we conclude that is constant, a contradiction.

One thought on “Fundamental Theorem of Algebra (With Picard’s Little Theorem)”

“we see that the sequence of z_k converges to some z”. Isn’t it that z_k isn’t necessarily convergent but that it has a convergent subsequence by the Bolzanno Weirstrass theorem? Also, since we know no such sequence z_k can exist then there are only finitely many z_k and thus there are plenty of points omitted from the set of interest.

“we see that the sequence of z_k converges to some z”. Isn’t it that z_k isn’t necessarily convergent but that it has a convergent subsequence by the Bolzanno Weirstrass theorem? Also, since we know no such sequence z_k can exist then there are only finitely many z_k and thus there are plenty of points omitted from the set of interest.